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Math Help - Trig Help... exsecant?

  1. #1
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    Trig Help... exsecant?

    i was wondering if anyone could lend me a little trig help... given the radius and theta, what would the two lengths indicated by question marks be? exsecant? i'm trying to program some animation, but my trig skills are nonexistent... thanks for your help.


    link to diagram...
    http://wilbcorp.com/links/trig.jpg

    -wilbur
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  2. #2
    MHF Contributor
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    Quote Originally Posted by cadete_b7 View Post
    i was wondering if anyone could lend me a little trig help... given the radius and theta, what would the two lengths indicated by question marks be? exsecant? i'm trying to program some animation, but my trig skills are nonexistent... thanks for your help.


    link to diagram...
    http://wilbcorp.com/links/trig.jpg

    -wilbur
    Let us call the given central angle as theta.
    Let us call the given radius of the circle as r.
    Let call the upper "?" as x, and the lower "?" as y.
    Let us call the hypotenuse of the legs x and y as z.
    So the whole vertical line from the center of the circle up to the bottom of y is (r+z).

    In the large right triangle, whose hypotenuse is (r+z):
    cos(theta) = r / (r+z)
    So,
    (r+z)cos(theta) = r
    (r+z) = r / cos(theta) = r*sec(theta)
    z = r*sec(theta) -r
    z = r*[sec(theta) -1] -------**

    In the smaller right triangle, whose hypotenuse is z:
    Since r and y are parallel ...they are both perpendicular to the same green line...then the angle between y and z is equal to theta.
    Then,

    sin(theta) = x / z
    So,
    x = z*sin(theta)
    x = r[sec(theta) -1]*sin(theta) ------------answer.

    cos(theta) = y / z
    y = z*cos(theta)
    y = r[sec(theta) -1]*cos(theta) --------answer.

    ------------------
    EDIT:

    Or, x = r[1 /cos(theta) -1] sin(theta) = r[sin(theta)/cos(theta) -sin(theta)] = r[tan(theta) -sin(theta)]

    And, y = r[cos(theta) /cos(theta) -cos(theta)] = r[1 -cos(theta)].
    Last edited by ticbol; July 26th 2008 at 10:51 PM.
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  3. #3
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    thanks.
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